Question

A boy of 50 kg mass is running with a velocity of 20 m/s. He jumps over a stationary cart of 2 kg while running. Find the velocity of cart after jumping of boy. (1 Point)​

Answer 1

Answer:

Given that:

• Mass of body A = 50 kg (boy)
• Initial velocity of boy = 20 mps
• Mass of body B = 2 kg (cart)
• Initial velocity of cart = 0

To calculate:

• Combined velocity of system

Solution: Combined velocity of system = 19.23 mps.

Explanation:

Using concept:

Conservation of law of momentum

Using formula:

${\small{\underline{\boxed{\sf{\rightarrow \: m_A u_A + m_B u_B \: = m_A v_A + m_B v_B}}}}}$

(Where, ${\sf{m_A}}$ denotes mass of object one, ${\sf{u_A}}$ denotes initial velocity of object one, ${\sf{m_B}}$ denotes mass of object two, ${\sf{u_B}}$ denotes initial velocity of object two, ${\sf{v_A}}$ denotes final velocity of object one, ${\sf{v_B}}$ denotes final velocity of object two.)

Required solution:

$:\implies \sf m_A u_A + m_B u_B \: = m_A v_A + m_B v_B \\ \\ :\implies \sf 50(20) + 2(0) = (50+2)v \\ \\ :\implies \sf 1000 + 0 = 52v \\ \\ :\implies \sf 1000 = 52v \\ \\ :\implies \sf 1000/52 = v \\ \\ :\implies \sf v \: = 19.23 \: ms^{-1}$